Question

1. Block A (m 4.5 kg) has a massless spring attached to its side. Block B (m 1.1 kg) is squeezed against the spring compressing it 0.12 m. The blocks are held this way at rest for a short period of time. The blocks are then released and explode away from each other. Block B flies east at 3.8 m/s. a. Calculate the speed and direction Block A travels after the release. b. Calculate the kinetic energy of block A after the explosion. c. Calculate the kinetic energy of block B after the explosion. d. What is the total kinetic energy of both blocks after the explosion e. How much energy was stored in the spring when it was compressed? f. What is the spring constant, k, of the spring?

Answer 1

(a) using conservation of momentum

Initially both blocks were at rest,

so,

m1v1f + m2v2f = 0

m1v1f = - m2v2f

v1f = -m2v2f / m1

v1f = - 1.1 * 3.8 / 4.5

v1f = - 0.9288 m/s

negative sign indicates that block A moves to the west (-x direction)

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K.E of block A = 1.2mv² = 0.5*4.5*0.9288² = 1.941 J

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K.E of block B = 0.5*1.1*3.8² = 7.942 J

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Total kinetic energy = 9.883 J

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As per energy conservation, the energy stored when compresses should be 9.883 J

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1/2kx² = 9.883

k = 2*9.883 / 0.12²

k = 1372.63 N/m